Question

find the value of a and b if (3+√2)/(3-√2)=a+b√2​

Answer 1

Answer:

$\mathrm{solve\:for\:a,}\:\frac{\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)}=a+b\sqrt{2}\quad :\quad a=\frac{11+6\sqrt{2}}{7}-\sqrt{2}b$

Step-by-step explanation:

$\frac{\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)}=a+b\sqrt{2}$

$\mathrm{Switch\:sides}$

$a+b\sqrt{2}=\frac{3+\sqrt{2}}{3-\sqrt{2}}$

$a+b\sqrt{2}=\frac{3+\sqrt{2}}{3-\sqrt{2}}$

$\mathrm{Subtract\:}b\sqrt{2}\mathrm{\:from\:both\:sides}$

$a+b\sqrt{2}-b\sqrt{2}=\frac{3+\sqrt{2}}{3-\sqrt{2}}-b\sqrt{2}$

$a=\frac{11+6\sqrt{2}}{7}-\sqrt{2}b$